Question

An object is shot upwards, from the ground,

with an iniítial velocity of 120 m/s. How high
will it be after 4.0s?


​

Answer 1

Answer:-

• Distance covered by the object after 4 sec = 401.6 metres

Given:-

• [The Object is shot upwards, therefore g will be negative]
• Acceleration due to gravity, a = g = -9.8 m/s²
• Initial velocity of object, u = 120 m/s
• Time for which it travelled upward, t = 4.0 sec

To find:-

• Height/distance covered by stone, s =?

Formula required;-

• The Second equation of motion
•         s = u t + 1/2 a t²

[ Where s is distance covered, u is initial velocity, t is time taken and a is acceleration ]

Calculation:-

Using second equation of motion

• → s = u t + 1/2 a t²
• → s = ( 120 ) · ( 4 ) + 1/2 · ( -9.8 ) ( 4 )²
• → s = 480 - 78.4  
• → s = 401.6 m

Therefore,

• Object will cover a distance of 401.6 metres after 4.0 seconds.

Answer 2

Answer:

Therefore, Object will cover a distance of 401.6 metres after 4.0 seconds

Explanation:

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